Is stopping something work?

If somebody pushes against a mass moving with $3 \frac{m}{s}$ to slow it down to $2 \frac{m}{s}$, he will drain the moving system of kinetic energy. Does he do work then or does he consume work?

My homework problem is that somebody is on a train, splits the cars and pushes one away making all other ones slower. The total energy of the system is lower, but he did some work obviously since he pushed it.

So do I add the two changes in kinetic energy (which yields a negative number) or do I add their absolute values?

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If you actually gave us your complete homework problem, then your homework problem is wrong; it describes an impossible occurrence.

Assuming all the cars are the same weight, if you push one car away from the other three, in order to maintain conservation of momentum, the change in velocity of the lead car should be three times that of the change in velocity of the others. That is, the forward car could now be going at 6 m/s and the other three at 2 m/s. The kinetic energy before the push is then proportional to $$\frac{1}{2} \cdot 4 \cdot 3^2 = 18.$$ The kinetic energy after the push is now proportional to $$\frac{1}{2}\left( 3 \cdot 2^2 + 1 \cdot 6^2 \right) = 24,$$ an increase.

Your intuition that you need to do work to push the car away is completely correct.

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The increase in total kinetic energy isn't a problem since this is provided by the person acting as a spring. The major problem is conservation of momentum cannot be conserved with these numbers, as you pointed out, no matter what mass you give the person. –  Larry Harson Nov 19 '11 at 20:06

Work is signed. When you apply a force that slows something down, you do negative work on it.

With the train example, you would not use the absolute values. The work done would be positive on cars that get faster and negative on cars that get slower.

To tell whether the work will be positive or negative, use the formula

$$\mathrm{d}W = \vec{F}\cdot \mathrm{d}\vec{x}$$

If the angle between the force and displacement is acute, the work is positive. If it is right, the work is zero. If it's obtuse, the work is negative.

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Okay, the person on the train pushes the cars apart. From the person's view (CMS), he does positive work in either direction since both cars move away from his old position. But from the rails he does work against the trailing cars which move against the force … so it would be negative. Since my POW are the rails, he does negative work on the trailing cars and therefore a negative total amount? –  queueoverflow Nov 19 '11 at 10:27
I don't know what POW is, but yes, I think you're saying it right. Work is not invariant - if you look at it in different reference frames it will be different. By the way, the problem you're writing about isn't a very good one. It doesn't conserve momentum so there have to be other forces around, but they aren't specified. –  Mark Eichenlaub Nov 19 '11 at 12:21
I meant Point of View (POV). Sorry. And it does not conserve the momentum since I am failed to solve the right equation. With the first velocity being 2.5, it will conserve the momentum though. Thanks for pointing that out! –  queueoverflow Nov 19 '11 at 12:39
And then, the overall Energy of this system increases from the point of view of the rails. This makes more sense after all. –  queueoverflow Nov 19 '11 at 12:41
Would the downvoter care to explain? –  Mark Eichenlaub Nov 19 '11 at 14:41